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Dear GAP-Forum, and Stas,

as an alternative to Sven's solution (which will almost surely come up

with a matrix of a prescribed weight; I don't know if that's what you

want?), you could `flip an unfair coin' for every entry:

r:=[1..10]; p:=2; coin:=function() if Random(r)>p then return 0*Z(2); else return Z(2); fi; end; n1:=5; n2:=5; A:=List([1..n1],i->List([1..n2],j->coin()));

This gives a matrix in which on average 20% of the entries are ones.

This is probably less efficient than Sven's solution, but the underlying

probability model is easier. So it depends on your application which

solution to use.

Best regards,

Jan

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